How are Microgrid Networks Optimized to Prevent Voltage and Frequency Droop?
Short Answer ---- With the advent of multiple small generators and energy storage devices in the modern power network, innovative solutions to the unique problems and opportunities of these new systems must be found. One of these problems (or opportunities) is the issue of how to correct voltage and frequency droop in distributed Microgrids. This paper describes some of the methods that can be used to help mitigate this issue.B. Parr (2009, Apr. 29). "Swine flu's tweet tweet causes online flutter," Business Standard Online. Available: http://mashabble.com/2009/04/04/twitter-trends Smartgrids are collections of various power producing, storage, and distribution systems that are controlled in a manner to optimize the system in a number of different ways. Micorgrids are smaller potentially self-contained systems that may consist of Distributed Generators (DGs), storage units, and local distribution lines. Microgrids interface with the Maingrid through a Point of Common Connection (PCC). The DGs within the Microgrid may consist of Photovoltaic (PV) cells, wind generators, small fuel burning units, or other power producing units. The energy storage units may consist of Uninterruptable Power Supplies (UPSs), capacitor banks, batteries, or electrical cars connected to the power network. Within Microgrids, the distribution system is usually transmitted at a relatively low voltage in comparison to the voltages that can be transmitted throughout the Maingrid. In the case that there is a fault or generation failure in the Maingrid, Microgrids have the capability to isolate themselves from the Maingrid and perhaps maintain power within them by a process referred to as Islanding. Loads within the system can come online or go offline and when this occurs, it will cause fluctuations within the power system. This is seen as changes in the voltage or frequency levels. These changes are referred to as droop. In order to counteract the droop introduced by the changing power requirements, a control system that has the ability to adjust the power provided by DGs and storage units may be implemented. The power in a power system consists of two types of power, Real power (P) that is consumed by resistive loads, and Reactive power (Q) which is consumed by inductive or capacitive loads. The sum of these two powers is called the Apparent Power (S). S=P+Q \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) If power is transmitted across a lossless line (one without a resistive load) the real and reactive power can be related to the voltages on either side of the line and to the power angle by (2) and (3). P=\frac{V_{1}V_{2}}{X}sin(\theta) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) Q=\frac{V_{2}}{X}(V_{2}-V_1cos(\theta)) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) Since the power angle \theta will be small, we can say that sin(\theta)\approx\theta and cos(\theta)\approx 1 . This allows (2) and (3) to be rewritten as (4) and (5). \theta\approx \frac{PX}{V_{1}V_{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) (V_{2}-V_{1})\approx \frac{QX}{V_{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5) From (4) it is seen that the real power P has a large influence on the power angle \theta . Since the frequency is dependent on the power angle \theta , the frequency droop can be controlled by the varying the real power. Likewise by (5) it is seen that the voltage difference between (V_{2}-V_{1}) is heavily influenced by the reactive power Q. So the voltage droop in a system can be controlled by changing the reactive power. The control equations for frequency and voltage droop are shown in (6) and (7). One can think of the base frequency and base voltage as the target values, and the system values as the values for the frequency and voltage at a particular time. f = f_{0} - k_{p} (P - P_{0}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6) V = V_{0} - k_{q} (Q - Q_{0}) \, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) Where f \, is the system frequency :: f_{0} \, is the base frequency :: k_{p} \, is the frequency droop control setting :: P \, is the active power of the unit :: P_{0} \, is the base active power of the unit :: V \, is the voltage at the measurement location :: V_{0} \, is the base voltage :: Q \, is the reactive power of the unit :: Q_{0} \, is the base reactive power of the unit :: k_{q} \, is the voltage droop control setting Within Microgrids, the various DGs which generally produce dc voltage, will interface through the power grid through inverters. Inverters are devices that convert dc voltage to ac voltage. Also, the energy storage devices will interface with the grid and either supply or absorb power from the grid. Theses storage devices will require an inverter when supplying power to the grid and an ac to dc converter when absorbing power from the grid. These inverters and converters can be controlled either locally by sensing the frequencies and voltages at their location, or controlled remotely by a Supervisory Control and Data Acquisition System (SCADA). A house with an example power profile over a 24 hour period is shown in Figure 2. The red line indicates the power that is being used by the house. The regular high power low power levels are indicative of the relatively high power use of an air conditioning unit cycling during the summer time. It is seen that the power usage spikes in early evening when the residents return home from the day. The light green line indicates the production of power from a PV array and the cyan line indicates the power output of a wind generator. The blue line is the grid power. The grid power (often hidden by the red line) is positive during the night and early morning. During the day, the excess power produced by the DG systems will at times provide power to the grid. This is shown when the grid power goes negative. The power is supplied to the grid through an inverter that is connected to the PCC of the house. Depending in the type of control system utilized, distributed local control or hierarchical centralized control; different modeling systems may be used. In general for distributed control using Nash game algorithm provides relatively good results, but in centralized control, the Stackleberg game algorithm provides superior results. Long Answer ---- Game Selection # Cooperative games are used where binding contracts are possible to reach a pareto-superior position for both players. (No fix power power output and no fixed load demand) # Non Cooperative Game: Non-cooperative games are used when negotiation and enforcement of contracts are not possible. (Independently adjusting power according to need). # Main types of Non Cooperative games * In non-cooperative games where players make their decisions simultaneously, the game is a Nash game. * Nash game is applied to capture stability notions. * In non-cooperative games where players make decisions sequentially, then the game is called a Stackelberg Game. '''Stackelberg game is important in networks with hierarchy. It states that if the leader knows the optimal response of the follower, then it can play the Stackelberg strategy to optimize its objective or performance index. '''Simple Stackelberg Model * One player (leader) has dominate influence over another. * Typically there are two stages: * One player moves first. * Then the other follows in the second stage. * Can be generalized to have multiple groups of players. * Static games in both stages. Main Theme Leader plays by backward induction, based on the anticipated behaviour of his/her follower. Game Solution In Figure 2, the predicted PV generation, load profiles and the current hour choices of β0; and Pμi are utilized; and the average values of the parameters for the upcoming hours are calculated and used to estimate the performance indices from hour (k+1) to N. Thus, the performance indices are modeled as: Jμ(βi(k), Pμ(k)) = β(k)Pμi(k) + (N - K)βiavg(k + 1→N)Pμiavg(k+1→N)). (8) Jt(βi(k), Pμ(k)) = aiPGi(k) + β(k)Pμi(k) + (N - K)+ 1→N)PGiavg(k+1→N)) + βiavg(k + 1→N)Pμiavg(k+1→N)). (9) To solve the [https://en.wikipedia.org/wiki/Game_theory game problem, the following steps are carried out to calculate performance indicies Jμ in (18) and Jt; in (19). 1. The active power Pμ(k) flowing from the microgrid (on the ith bus) to the main bus grid at k hour is given by (12) 2. Assume that the storage level of a microgrid should return to its initial value after N hours, then the average active power from hour (k+1) to N, Pμavg(k+1→N) is given by: Pμiavg(k+1→ N)= avgPμ(k+1→N) + Σ l=1kΔ Ei(l) N-K (10) where avgPμi(k+1→N) is the predicted average power from historical data without considering the storage. 3. Conventional generation PGi(k) (on the the ith bus) at hour k is given by: PGi(k) = PDi(k)-Pμi(k). (11) And the predicted average generation from hour(k+1) to N is given by: PGiavg(k+1→N) = PDavg(k+1→N)-Pμiavg(k+1→N) (12) where PDavg (k+1→N) is the predicted average demand from hour(K+1) to N. 4. Using PGi is computed by: β(k) = β0PGi(k) - PGi*(k) PGi*(k) (13) where PGi* is the optimal operation of the power of the conventional generator. β 0 is a base price and η i(k) is a variable, which the main grid perturbs to find different energy price offers to play the game. 5. Therefore, βiavg(k+1→N) = β0+ ηi(k) PGiavg(k+1→N) - PGi*(k) PGi*(k) (14) Hence, a matrix can be constructed as : 1. For each η i (k, l ) (k, j ) for j = 1,2,....M1 is corresponding to Pμi(k) can be found such that : Jμi [ ηi(k,j),Pμi(k)]is maximized. (15) and that Pμi(k) is denoted as PμiSηi(k,j). The Stackelberg solution of the main grid is ηi(k,l) for some l such that: Jt (k,l), PμiS (ηi (k, l ) ) ≤ Jt [ ηi (k, j ), PμiS (k, j ) ) ] To illustrate this algorithm, a simple example is provided as folows: for all j = 1,2,....,M1 Example 1. Suppose that the matrix game is shown Table 2. The Stackelberg solution with the main grid as the leader is obtained as follows: 1. For ηi(k)=1, Pμi(k) = 1 maximizes Jμi because Jμi(1,1) = 10 is greater than Jμi(10,8) = 8. For ηi(k)=1.2, Pμi(k) = 0.8 maximizes Jμi because Jμi(1.2,0.8) = 7 is greater than Jμ(1.2,1) = 6 2.As in Table 2 Jt(1.2,0.8) = 5 is less than Jt(1,1) = 9 the Stackelberg solution of the main grid is η(k) = 1.2 References ---- __FORCETOC__